Divergence of F = div $\bar{F} = \bar{V} \cdot \bar{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{1}}{\partial y} + \frac{\partial F_{1}}{\partial z}$

Given: 2

$\bar{F} = F_{1}\hat{i} + F_{2}\hat{j} + F_{3}\hat{k}$

$\therefore \frac{\partial F_{1}}{\partial x} = y[x(2e^{2x} + e^{2x}] \\ = y[2xe^{2x} + e^{2x}] \\ = ye^{2x}[2x + 1]\\ \frac{\partial F_{2}}{\partial y} = xcosz(2y) = 2xycosz \\ \frac{\partial F_{3}}{\partial y} = 0\\ \therefore \bar{V} \cdot \bar{F} = ye^{2x}(2x + 1) + 2xycosz$

Curl of F = $\bar{V} \times \bar{F} = \bigg(i\frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial }{\partial z} \bigg) \times F_{1}\hat{i} + F_{2}\hat{j} + F_{3}\hat{k}$

$\therefore \bar{V} \times \bar{F} = \hat{i}[x^{2} (-sinxy)(x) - xy^2 (-sinz)] + \hat{j}[x^2 (-sinxy)(y) + 2xcosxy - n] + \hat{k}[y^2 cosz - xe^{2x}]$

$\therefore \bar{V}\times \bar{F} = \hat{i}[xy^2 (sinz) - x^3(sinxy)] + \hat{j}[2x(cosxy) - x^2ysinxy] + \hat{k}[y^2 cosz - xe^{2x}]$